sinA=3/5thancosA=howmuch
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CosA=4/5
as SinA=perpendicular/hypotenuse
H²-P²=B²
25-9=B²
B=√16
B=4
CosA=Base/hypotenuse
CosA=4/5
as SinA=perpendicular/hypotenuse
H²-P²=B²
25-9=B²
B=√16
B=4
CosA=Base/hypotenuse
CosA=4/5
Answered by
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sinA=P/H=3/5
by using Pythagoras theorem,
b=√5 ka square-3 ka square
b=√25-9
b=√16
b=4
cosA=b/h=4/5
by using Pythagoras theorem,
b=√5 ka square-3 ka square
b=√25-9
b=√16
b=4
cosA=b/h=4/5
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